A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a)  total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^–5 m², and the free electron density in copper is given to be about 10^–29 m^–3.)

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Given,
Number of turns, N = 20 
Radius of the coil, r = 10 cm = 10 x 10^–2 m
Magnetic field, B = 0.10 T
Current carried, I = 5.0 A
θ = 0° (angle between field and normal to the coil)
Area of the coil, A = $\mathrm{\pi }$r² 
                          = $\mathrm{\pi }$ x (10 x 10^–2)²
                          = $\mathrm{\pi }$ x 10^–2 m²

(a) Torque, τ = NIBA sin θ
                   = 20 x 5.0 x 0.10 x $\mathrm{\pi }$ x 10^–2 sin 0°
                   = 20 x 5.0 x 0.10 x $\mathrm{\pi }$ x 10^–2 x 0 = 0

(b) Net force on a planer current loop in a magnetic field is always zero, as net force due to couple of force is zero. 

(c) If v_d is the drift velocity of electron

                       F = qv x B
                         = ev_d. B sin 90°

Force on one electron = Be  ${\mathrm{v}}_{\mathrm{d}} = \mathrm{Be}\frac{\mathrm{I}}{\mathrm{neA}} = \frac{\mathrm{BI}}{\mathrm{nA}}$
Here,  

                    $\mathrm{n} = {10}^{29} {\mathrm{m}}^{-3}, \mathrm{A} ={10}^{-5}{\mathrm{m}}^2$

$\therefore$ Force on one electron = $\frac{0.10 \times 5.0}{{10}^{29} \times {10}^{-5}} = 5 \times {10}^{-25}\mathrm{N}.$

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Given, a galvanometer coil.
Resistance on galvanometer, G = 15 Ω
Current pasing through the galvanometer,I_g = 4 mA = 4 x 10^–3 A
Current across the ammeter, I = 6 A 

Using formula, 
                       $\boxed{\mathrm{S} = \frac{{\mathrm{I}}_{\mathrm{g}} . \mathrm{G}}{\mathrm{I}-{\mathrm{I}}_{\mathrm{g}}}}$ 

Putting values,
                        $$\begin{gathered} \mathrm{S} = \frac{4 \times {10}^{-3} \times 15}{6-0.004} \\ = \frac{60 \times {10}^{-3}}{5.996} \\ = 10 \times {10}^{-3}\mathrm{\Omega } \end{gathered}$$ 

                         $\mathrm{S} = 10 \mathrm{m} \mathrm{\Omega }$ 

Therefore, a resistance of 10mΩ should be connected so as to convert the galvanometer into an ammeter.

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a) Diameter of cylindrical region
= 20 cm = 0.20 m
Clearly,    l = 0.20 m. Also, θ = 90°
F = BIl sin θ = 1.5 x 7 x 0.20 sin 90° N
= 2.1 N
Using Fleming's left-hand rule, we find that the force is directed vertically downwards.

(b)    If l₁ is the length of the wire in the magnetic field, then,
F₁ = BIl₁ sin 45°
But l₁ sin 45° = l
∴ F₁ = BIl = 1.5 x 7 x 0.20 N = 2.1 N
The force is directed vertically downwards by Fleming's left hand rule.
(c)    When the wire is lowered by 6 cm, the length of the wire in the cylindrical magnetic field is 2x.
Now,              
                  
              
                 
The force is directed vertically downwards.

The result is true for any angle between current and direction of . This is because I sin θ remains constant i.e., 20 cm.

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(a) Given, 
Magnetic field along the positive z-direction = 3000 G = 3000 x 10^–4 T = 0.3 T
Current carried by the loop, I = 12 A
Area of rectangular loop, A = 10 x 5 cm² 
                                      = 50 x 10^–4 m²

Torque on the loop is given by,
                  $\boxed{\mathrm{\tau } = \mathrm{BIA} \cos \mathrm{\theta }}$ 

where, θ is the angle between the plane of loop and direction of magnetic field.
Here, θ = 0° . 

Therefore, torque is, τ = 0.3 x 12 x 50 x 10^–4 
                                = 1.8 x 10^–2 Nm. 

Using Fleming's left hand rule we can say that the direction of torque is along negative y direction. 

(b) The torque acting is the same as in case (a) but, the direction of torque is along side 10 cm. 

(c) The magnitude of torque is equal to 1.8 x 10^–2 Nm along - x direction of torque on lower arm of 5 cm towards – y axis.
(d)    This case is similar to (c). Direction of torque is 60°.
(e)    zero. (∵ angle between plane of loop and direction of magnetic field is 90°)
(f)    zero. 

Force is zero in each case. Stable equilibrium is corresponded by case (e).

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Given, a galvanometer coil.
Resistance on the galvanometer, R = 12 Ω
Current on galvanometer, I_g =3 mA = 3 x 10^–3 A
Voltage ,V = 18 V

By using formula,

                      $\boxed{\mathrm{V} = {\mathrm{I}}_{\mathrm{g}}(\mathrm{R}+{\mathrm{R}}_{\mathrm{g}})}$

                     $\frac{\mathrm{V}}{{\mathrm{I}}_{\mathrm{g}}} = \mathrm{R}+{\mathrm{R}}_{\mathrm{g}}$ 
or,         
                      $$\begin{gathered} \mathrm{R} = \frac{\mathrm{V}}{{\mathrm{I}}_{\mathrm{g}}}-{\mathrm{R}}_{\mathrm{g}} \\ = \frac{18}{3\times {10}^{-3}}-12 \\ \mathrm{R} = 6 \times {10}^3-12 = 5988\mathrm{\Omega } \end{gathered}$$